Unique factorization of ideals in a quadratic field

Question

"Suppose $k = \mathbb{Q}(\sqrt{d})$ is a real quadratic field ($d > 1$ a square-free integer) with fundamental unit $\varepsilon$, normalized as usual so that $\varepsilon > 1$ with respect to the embedding of $k$ into $\mathbb{R}$ for which $\sqrt{d} > 0$. If $\operatorname{Norm}_{k/\mathbb{Q}}(\varepsilon) = +1$, then, by Hilbert’s Theorem 90, $$ \tag{1}\label{Hilbert90} \varepsilon = \sigma(\alpha)/\alpha $$ for some $\alpha \in \mathbb{Q}(\sqrt{d})$, which may be assumed to be an algebraic integer (for example, take $\alpha = \sigma(\varepsilon) + 1$). The principal ideal $(\alpha)$ is invariant under $\sigma$ (that is, is an ambiguous ideal) since $\sigma(\alpha)$ differs from $\alpha$ by a unit, so the element $\alpha$ satisfying \eqref{Hilbert90} can be further chosen so that the ideal $(\alpha)$ is the product of distinct ramified primes."

I got this from some paper.

Why $(\alpha)$ can be written as the product of distinct ramified primes?

Answer 1

It is not true that $(\alpha)$ is the product of distinct ramified primes. What is true is that there exists an $\alpha\in\mathcal{O}_k$ satisfying $(1)$ for which $(\alpha)$ is the product of distinct ramified primes.

The proof is simple. Take an $\alpha\in\mathcal{O}_k$ satisfying (1). Then $\sigma$ leaves the ideal $(\alpha)$ invariant, whence in its prime ideal factorization the split prime ideals can be grouped into $\sigma$-conjugate pairs. So the product of all split prime ideals occurring in $(\alpha)$ is generated by a positive rational integer. The same is trivially true for the product of all inert prime ideals occurring in $(\alpha)$. Finally, observe that the square of each ramified ideal is also generated by a positive rational integer. It follows that $(\alpha)$ can be written as $(n)\mathfrak{p}_1\dotsb\mathfrak{p}_m$, where $n$ is a positive rational integer, and each $\mathfrak{p}_j$ is a ramified prime ideal. Hence $\alpha':=\alpha/n\in\mathcal{O}_k$ satisfies $(1)$, because $\alpha$ satisfies $(1)$ and $\sigma(n)=n$, and also $(\alpha')=\mathfrak{p}_1\dotsb\mathfrak{p}_m$. Done.